What causes cherenkov radiation?

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    Cherenkov Radiation
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SUMMARY

Cherenkov radiation occurs when a charged particle, such as an electron, travels faster than the speed of light in a given medium, emitting light as a result. The semi-classical derivation presented in Schiff's "Quantum Mechanics" (2nd Edition, pages 267-271) provides a detailed explanation of the classical electric and magnetic fields involved, leading to the calculation of the number of photons emitted per unit path length. Specifically, the formula for the number of quanta radiated is given by dN = (1/137)(1 - (c²/n²v²))(dω/c) for finite indices of refraction, and simplifies to dN = (1/137)(dω/c) for an infinite index of refraction. This indicates that the total number of emitted quanta is dependent on the frequency interval dω.

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blokeice
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If you get an electron going faster than the local speed of light i know that it starts emitting light, but why does it do this?, furthermore, In a theoretical material with an infinite index of defraction (a.k.a light is stopped within it), wouldn't that cause the system to quickly go to near absolute zero?
 
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The best derivation of Cerenkov radiation by a fast charged particle I have seen is the semi-classical derivation in Schilff Quantum Mechanics (2nd Edition) pages 267-271. Schiff derives the classical E and H fields, and the resulting Poynting vector P = E X H. He then gets the number of quanta radiated per unit path length in frequency interval ω to ω+dω:
dN=\frac{1}{137}\left(1-\frac{c^2}{n^2v^2} \right)\frac{d\omega}{c} \text{ photons per unit length} which becomes for infinite index of refraction
dN=\frac{1}{137}\frac{d\omega}{c} \text{ photons per unit length}
So the total number of quanta depends on wnat interval the frequency interval dω covers. Normally, the index of refraction is n ≤1 for wavelengths less than ~ 1000 Angstroms (UV).
 

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